Question

Let $A B C D$ be a parallelogram. If
$A B =\hat{ i }+3 \hat{ j }+7 \hat{ k }, A D =2 \hat{ i }+3 \hat{ j }-5 \hat{ k }$ and $p$ is a unit vector parallel to $A C$, then $p$ is equal to

• A. $\frac{1}{3}\left(2\hat{i}+\hat{j}+2\hat{k}\right)$
• B. $\frac{1}{3}\left(2\hat{i}-2\hat{j}+\hat{k}\right)$
• C. $\frac{1}{7}\left(3\hat{i}+6\hat{j}+2\hat{k}\right)$
• D. $\frac{1}{7}\left(6\hat{i}+2\hat{j}+3\hat{k}\right)$
• E. $\frac{1}{7}\left(6\hat{i}+2\hat{j}-3\hat{k}\right)$

Answer 1

Answer: (C)

$ AC = AB + AD $
$=(\hat{ i }+3 \hat{ j }+7 \hat{ k })+(2 \hat{ i }+3 \hat{ j }-5 \hat{ k }) $
$=3 \hat{ i }+6 \hat{ j }+2 \hat{ k } $

Now, $p$ is a unit vector parallel to AC
Then, $ p =\frac{ AC }{| AC |} $
$=\frac{3 \hat{ i }+6 \hat{ j }+2 \hat{ k }}{\sqrt{9+36+4}} $
$\Rightarrow p =\frac{1}{7}(3 \hat{ i }+6 \hat{ j }+2 \hat{ k })$